Question

can anyone help me? i'm confused:

csc^2Atan^2A - 1 = tan^2A

Answer 1

\[\csc^2A = \frac{ 1 }{ \sin^2A }\]
\[\tan^2A = \frac{ \sin^2A }{ \cos^2A }\]
\[LHS = \csc^2A \tan^2A - 1 = \frac{ 1 }{ \sin^2A } \frac{ \sin^2A }{ \cos^2A } - 1= \frac{ 1 }{ \cos^2A } - 1 = \frac{ 1-\cos^2A }{ \cos^2A }\]
\[= \frac{ \sin^2A }{ \cos^2A } = \tan^2A = \]

alternatively from 1/cos^2A - 1 can use the Pythagorean trig identity sec^2A - 1 = tan^2A
\[\frac{ 1 }{ \cos^2A } - 1 \]

Answer 2

do u have to find A in this?

Answer 3

no.. just proving it

Answer 4

or just prove that this is right? that LHS is equal to RHS